burnbird16 wrote:I have a circle, centre (-3,2) and radius of sqr root 10. The question is asking me to find the equation of the tangent at (-2,5)....

What is the slope of the line containing the center and the point of tangency?

What is a basic property of tangents? That is, at what sort of "angle" does a tangent touch a circle? So what is the relationship between the line through the center and the tangency point, and the tangent line?

Then what is the slope of the tangent line?

Then what is the equation of the tangent?

burnbird16 wrote:A circle is tangent to the y-axis at y=3 and has one x-intercept at x=1. Determine the other x-intercept.

Since the circle is perpendicular to the y-axis, what is the y-coordinate of the center?

You have the equation (x - h)2 + (y - 3)2 = r2. You have the points (0, 3) and (1, 0) on the circle. Plug these into the equation. See where the resulting equations lead you....

stapel_eliz wrote:What is the slope of the line containing the center and the point of tangency?

What is a basic property of tangents? That is, at what sort of "angle" does a tangent touch a circle? So what is the relationship between the line through the center and the tangency point, and the tangent line?

Then what is the slope of the tangent line?

Then what is the equation of the tangent?

The slope of the first line is 3. I know that the basic property of tangents is that it will be perpendicular to the slope of its radius, so it touches the circle at a ninety degree angle, I think? I don't know.

The slope of the tangent line must be -1/3 x.

Therefore, the equation must be y=-1/3x + 13/3. Tell me if I'm on mark.